Question

1. Given a binary code, determine the number of errors that it can detect and the number of errors that it can correct. WRITE A PROGRAM PYTHON or JAVA. both would be nice

Answer 1

package Networking;

import java.util.*;

public class ErrorDectCorrect

{

static int setParityBit(int data[])

{

// Initializing count to zero which will count the number of 1.

int counter = 0;  

// Loops till data length

for(int c = 0; c < data.length; ++c)

// Increase count if value in array "data" is 1.

if(data[c] == 1)

++counter;

// Returns 0 if counter value is even number (even number of 1s)

if((counter % 2) == 0)

return 0;

// Otherwise returns 1 for odd number of 1

else

return 1;

}// End of method

// main method definition

public static void main(String args[])

{

// Scanner class object created

Scanner keyboard = new Scanner(System.in);

System.out.print("\n Error detection and correction using EVEN parity");

// Sets number of data bits

int numberOfDataBits = 4;

// Accepts 4 bits of data

System.out.print("\n\n Enter 4 data bits: ");

// Declares an array to accept data bits

int dataBits[] = new int[4];

for(int pos = numberOfDataBits - 1; pos >= 0; --pos)

{

System.out.print("Enter the value of D" + (pos + 1) + " -> ");

dataBits[pos]=keyboard.nextInt();

}// End of for loop

// Formula (2 ^ numberOfParityBits >= numberOfDataBits + numberOfParityBits + 1)

// Initializing it to zero.

int numberOfParityBits = 0;

// Calculating the value of k(number of parity bits).

while(Math.pow(2, numberOfParityBits) < (numberOfDataBits + numberOfParityBits + 1))

++numberOfParityBits;

System.out.println("\n" + numberOfParityBits +

" parity bits are required for the " +

"transmission of data bits.");

// Declares an array to store parity bits

int parityBits[] = new int[numberOfParityBits];

// Array to store the hamming code.

// (n + numberOfParityBits + 1) as we start from pos 1.

int hammingCode[] = new int[numberOfDataBits + numberOfParityBits + 1];

// Initialize each hammingCode array element to -1

for(int pos = 0; pos < 7; ++pos)

hammingCode[pos] = -1;

int count = 0;

int c = 2;

while(count < 4)

{

++c;

if(c == 4)

continue;

hammingCode[c] = dataBits[count];

++count;

}// End of while loop

int p1[] = {hammingCode[1], hammingCode[3], hammingCode[5], hammingCode[7]};

int p2[] = {hammingCode[2], hammingCode[3], hammingCode[6], hammingCode[7]};

int p3[] = {hammingCode[4], hammingCode[5], hammingCode[6], hammingCode[7]};

int parity[] = new int[3];

// Calls the method to set the value of parity bit

parity[0]= setParityBit(p1);

parity[1]= setParityBit(p2);

parity[2]= setParityBit(p3);

// Inserting the parity bits in the hamming code

hammingCode[1] = parity[0];

hammingCode[2] = parity[1];

hammingCode[4] = parity[2];

System.out.println("\n\n SENDER SIDE:");

System.out.print("\n The data bits entered are: ");

for(int pos = 3; pos >= 0; --pos)

System.out.print(dataBits[pos] +" ");

System.out.println("\n\n The Parity bits are: ");

for(int pos = 2; pos >= 0; --pos)

System.out.println("Value of P " + (pos + 1) + " is " + parity[pos] + " ");

System.out.print("\n\n The Hamming code is as follows :-" +

"\nD4 D3 D2 P3 D1 P2 P1 \n");

for(int pos = (numberOfDataBits + numberOfParityBits); pos > 0; --pos)

System.out.print(hammingCode[pos] + " ");

System.out.println();

System.out.println("\n Enter the hamming code with error at " +

"any position of your choice." +

"\nNOTE: ENTER A SPACE AFTER EVERY BIT POSITION." +

"\nError should be present only at one bit position");

for(int pos = 7; pos > 0; --pos)

hammingCode[pos] = keyboard.nextInt();

int p4[]={hammingCode[1], hammingCode[3], hammingCode[5], hammingCode[7]};

int p5[]={hammingCode[2], hammingCode[3], hammingCode[6], hammingCode[7]};

int p6[]={hammingCode[4], hammingCode[5], hammingCode[6], hammingCode[7]};

parity[0]= setParityBit(p4);

parity[1]= setParityBit(p5);

parity[2]= setParityBit(p6);

int position = (int)(parity[2] * Math.pow(2,2) +

parity[1] * Math.pow(2,1)+

parity[0] * Math.pow(2,0));

System.out.print("\n\n RECEIVER:");

System.out.print("\n Error is detected at position " + position +

" at the receiving end.");

System.out.print("\n Correcting the error.... ");

if(hammingCode[position] == 1)

hammingCode[position]=0;

else

hammingCode[position] = 1;

System.out.print("\n The correct code is ");

for(int pos = 7; pos > 0; --pos)

System.out.print(hammingCode[pos] + " ");

}// End of main method

}// End of class

Sample Output:

Error detection and correction using EVEN parity

Enter 4 data bits: Enter the value of D4 -> 1
Enter the value of D3 -> 0
Enter the value of D2 -> 1
Enter the value of D1 -> 0

3 parity bits are required for the transmission of data bits.

SENDER SIDE:

The data bits entered are: 1 0 1 0

The Parity bits are:
Value of P 3 is 0
Value of P 2 is 1
Value of P 1 is 0

The Hamming code is as follows :-
D4 D3 D2 P3 D1 P2 P1
1 0 1 0 0 1 0

Enter the hamming code with error at any position of your choice.
NOTE: ENTER A SPACE AFTER EVERY BIT POSITION.
Error should be present only at one bit position
1 0 1 1 0 0 1

RECEIVER:
Error is detected at position 7 at the receiving end.
Correcting the error....
The correct code is 0 0 1 1 0 0 1